$f(x) = -6x^{2}+4(h(x))$ $h(x) = -3x$ $ f(h(1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $h(1)$ . Then we'll know what to plug into the outer function. $h(1) = (-3)(1)$ $h(1) = -3$ Now we know that $h(1) = -3$ . Let's solve for $f(h(1))$ , which is $f(-3)$ $f(-3) = -6(-3)^{2}+4(h(-3))$ To solve for the value of $f$ , we need to solve for the value of $h(-3)$ $h(-3) = (-3)(-3)$ $h(-3) = 9$ That means $f(-3) = -6(-3)^{2}+(4)(9)$ $f(-3) = -18$